Linear Equations and Inequalities: Question 5

Worked solution

Part (a): solving the equation with algebraic fractions

$\frac{x+3}{4} - \frac{x-2}{3} = 1$

Step 1: Clear the denominators. The lowest common multiple of $4$ and $3$ is $12$, so multiply every term on both sides by $12$:

$12 \times \frac{x+3}{4} - 12 \times \frac{x-2}{3} = 12 \times 1$

$3(x+3) - 4(x-2) = 12$

Step 2: Expand each bracket carefully, watching the signs:

$3(x+3) = 3x + 9, \qquad -4(x-2) = -4x + 8 \quad (\text{since } -4 \times -2 = +8)$

So the equation becomes:

$3x + 9 - 4x + 8 = 12$

Step 3: Simplify the left-hand side.

$(3x - 4x) + (9 + 8) = 12$

$-x + 17 = 12$

Step 4: Solve for $x$.

$-x = 12 - 17 = -5$

$x = 5$

$\boxed{x = 5}$

Check: $\dfrac{5+3}{4} - \dfrac{5-2}{3} = \dfrac{8}{4} - \dfrac{3}{3} = 2 - 1 = 1$ ✓

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Part (b): evaluating $2x - 7$

Using $x = 5$:

$2x - 7 = 2(5) - 7 = 10 - 7 = 3$

$\boxed{2x - 7 = 3}$

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Final answers

• (a) $x = 5$
• (b) $2x - 7 = 3$